2025年新高考1卷第11题, 从条件简单化简即得 A 选项, 然而直接想要证明 $C = \frac{\pi}2$ 并不省事.
为了方便计算, 令 $\alpha = \frac{A+B}2$, $\beta = \frac{A-B}2$, 则 $A = \alpha + \beta$, $B = \alpha – \beta$.
由题可知 \begin{align*} \sin C &= \sin^2 A + \sin^2 B \\ \sin A \cos B + \cos A \sin B &= \sin^2 A + \sin^2 B \\ \sin A ( \sin (\alpha + \beta) – \cos (\alpha – \beta)) &= \sin B ( \sin(\alpha – \beta) – \cos (\alpha + \beta) ) \\ \sin A (\sin \alpha – \cos \alpha) (\sin \beta – \cos \beta) &= \sin B (\sin \alpha – \cos \alpha) (\sin \beta + \cos \beta) \end{align*}
若 $\sin \alpha – \cos \alpha = 0 \implies A+B = \frac{\pi}2 \implies C = \frac{\pi}2$.
接下来继续处理剩余部分. \begin{align*} \sin (\alpha + \beta) (\sin \beta – \cos \beta) &= \sin (\alpha – \beta) (\sin \beta + \cos \beta) \\ \sin \alpha \cos\beta \sin\beta + \cos\alpha \sin^2 \beta &- \sin\alpha \cos^2 \beta – \cos \alpha \cos\beta \sin \beta \\ = \sin \alpha \cos\beta \sin\beta – \cos\alpha \sin^2 \beta &- \sin\alpha \cos^2 \beta + \cos \alpha \cos\beta \sin \beta \\ \cos \alpha \sin \beta (\sin \beta – \cos \beta) &= 0 \end{align*}
同理 $\sin \beta = 0 \implies A=B \implies \sin A = \cos A \implies C = \frac{\pi}2$.
当然, 需要注意到 $\cos \alpha = 0 \implies A+B = \pi$; $\sin\beta – \cos\beta = 0 \implies A-B = \frac{\pi}2 \implies \sin^2 B = 0$. 均矛盾!